A light string passes over a frictionless pulley. To one of its ends a mass of 6 kg is attached. To its other end a mass of 10 kg is attached. The tension (in newton) in the thread will be [Take g = 9.8 m/s2]

Question

A light string passes over a frictionless pulley. To one of its ends a mass of 6 kg is attached. To its other end a mass of 10 kg is attached. The tension (in newton) in the thread will be [Take g = 9.8 m/s2] <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/08edc4f44299e24dbce037729585932c-.png" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

T= (2m1m2/m1 +m2)g = (2 × 10× 6/10 +6) ×9.8 = 73.5N

Q. A light string passes over a frictionless pulley. To one of its ends a mass of $6$ kg is attached. To its other end a mass of $10$ kg is attached. The tension (in newton) in the thread will be [Take $g = 9.8$ m/s $^{2}$ ]

$T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$
$= \frac{2 \times 10 \times 6}{10 + 6} \times 9.8 = 73.5 N$

Q. A light string passes over a frictionless pulley. To one of its ends a mass of 6 kg is attached. To its other end a mass of 10 kg is attached. The tension (in newton) in the thread will be [Take g=9.8 m/s2]

T=m1​+m2​2m1​m2​​g =10+62×10×6​×9.8=73.5N

Q. A light string passes over a frictionless pulley. To one of its ends a mass of $6$ kg is attached. To its other end a mass of $10$ kg is attached. The tension (in newton) in the thread will be [Take $g = 9.8$ m/s $^{2}$ ]

$T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$
$= \frac{2 \times 10 \times 6}{10 + 6} \times 9.8 = 73.5 N$